explicit formula.

[samus101]

Can someone please explain to me how to find the explicit formula for P???

1/4ln(P-2)-1/4ln(P+2) = 1/3t^3-4t+C

The answer is

P = 2 (1+Aexp(4/3t^3-16t)) / (1-Aexp(4/3t^3-16t))

I'm just not to sure how to get to the answer. Thanks.

 

[tkhunny]

Properties of Logarithms, for suitable a, b, and c.

\(\displaystyle a\cdot log(b) = log\left(b^{a}\right)\)

Do that twice.

\(\displaystyle log(a) - log(b) = log\left(\frac{a}{b}\right)\)

Do that once.

\(\displaystyle log_{b}(a) = c \iff b^{c} = a\)

And you are nearly done, except for a little algebra.

I am a little curious where "A" came from.

 

[HallsofIvy]

Can someone please explain to me how to find the explicit formula for P???

1/4ln(P-2)-1/4ln(P+2) = 1/3t^3-4t+C

The answer is

P = 2 (1+Aexp(4/3t^3-16t)) / (1-Aexp(4/3t^3-16t))

I'm just not to sure how to get to the answer. Thanks.

What you give as "the answer" is wrong because there is no "C".

ln(a)- ln(b)= ln(a/b) so your first equation is
\(\displaystyle (1/4)ln(\frac{P-2}{P+2})= (1/3)t^3- 4t+ C\)
Now multiply both sides by 4:
\(\displaystyle ln(\frac{P-2}{P+2})= (4/3)t^3- 16t+ 4C\)

If you are expected to do a problem like this then you are expected to know that "ln(x)" and "exp(x)" are inverse function- ln(exp(x))= exp(ln(x))= x. So taking the exponential of both sides of that equation gives
\(\displaystyle \frac{P-2}{P+2}= exp((4/3)t^3- 16t+ 4C)\).

Now we can simplify the writing a little by writing \(\displaystyle A= exp((4/3)t^3- 16t+ 4C\):
\(\displaystyle \frac{P-2}{P+2}= A\)
multiplying both sides by P+ 2, \(\displaystyle P- 2= A(P+2)= AP+ 2A\)
\(\displaystyle P- AP= 2A+ 2\)
\(\displaystyle (1- A)P= 2(A+ 1)\)
\(\displaystyle P= \frac{2(1+ A)}{A- 1}\)

And just replace "A" with \(\displaystyle exp((4/3)t^3- 16t+ 4C\).